This is really three equations for each of the components of $\myv E$, like
$$\left[\frac{\del^2}{\del x^2} +\frac{\del^2}{\del y^2} +\frac{\del^2}{\del z^2}\right]E_x(x,y,z,t)=\mu\epsilon\frac{\del^2}{\del t^2}E_x(x,y,z,t).$$

So, $E_x$, $E_y$, and $E_z$ each obey an equation which actually turns out to be a wave equation:
$$ \nabla^2 u
-\frac{1}{v^2}\frac{\del^2 u}{\del t^2}=0,$$

Figure out what $k$ has to be (in terms of $\omega$ and $v$) in order for this to be a solution to the wave equation:
$$u(x,y,z,t) = \cos(kz-\omega t).$$
...and what direction is this solution moving in? [WA picture of $u(x,y,z,t=0)$ and at $t=0.5$ with $v=1$, $\omega=1$].

What is happening in the $x$- and $y$-directions?

Here's a conceptual picture of the function $\cos(kz-\omega
t)$. The planes connect all the points in space at one instant in time
where $u=1$ (the peak value of the sinusoidally oscillating cosine function).
These planes move in the $z$-direction with speed $v$. This is a called a 'plane
wave' solution.

Come up with a $u(x,y,z,t)$ that is moving in the $-\uv y$ direction.

We could write these last two solutions as...$$u=\cos(k\uv z\cdot \myv r - \omega t)$$
and
$$u=\cos(k(-\uv y)\cdot \myv r - \omega t)$$
with $k=\omega/v$ in each case.

Come up with a $u(x,y,z,t)$ that is moving in the direction which is half-way between $\uv x$ + $\uv y$.

Plane wave solution

A very general solution to the wave equation for any one of the components of the electric field, is a plane wave solution travelling with speed $v$ in the direction of $\myv k$:
$$ u=C\cos\left(\myv k \cdot \myv r+\omega t + \delta\right).$$
The magnitude of the 'wave vector' $\myv k$ is:
$$|\myv k| \equiv k =\frac{\omega}{v}=\sqrt{\mu\epsilon} \omega$$

The speed in free space is (in the SI units we've been using, this should come out in m/s)...
$$v_0 = \frac{1}{\sqrt{\mu_0\epsilon_0}}=\frac{1}{\sqrt{4\pi\times 10^{-7}\cdot8.85\times 10^{-12}}}. $$

In optics, the index of refraction was defined as the speed of light in vacuum divided by the effective speed of light in a medium:
$$n=c/v = \frac{\sqrt{\mu\epsilon}}{\sqrt{\mu_0\epsilon_0}}.$$
(For most dielectrics, $\mu\approx \mu_0$...)

The argument of the cosine function is in radians, so the 'angular frequency' $\omega$ apparently has units of [radians/sec]. Since there are $2\pi$ radians in one cycle, the frequency, $f$, of the wave (which has units of [cycles/sec]) is:
$$f=\frac{\omega}{2\pi}.$$

The wave vector has units of [radians/meter]. The wavelength, $\lambda$, of a wave can be thought of as having units of [meters/cycle], so
$$\frac{1}{\lambda}=\frac{k}{2\pi}.$$

The relation above between $k$ and $\omega$ becomes
$$v=\frac{\omega}{k}=\frac{2\pi f}{2\pi/\lambda}=f\lambda.$$

This is the equation for just one of the Cartesian components of $\myv E$. We can write all three amplitudes together as...
$$\myv E (\myv r, t) = Re\left[\myv{\mathcal E} e^{i\myv k \cdot \myv r -i\omega t}\right]$$
where $\myv{\mathcal E}$ is constant vector with 3 complex components.

If we had started with $\myv\grad\times(\myv\grad\times\myv B)$ instead, we would get identical wave equations for $\myv B$, and so a very general solution for $\myv B(x,y,z,t)$ is...
$$\myv B(\myv r, t) = Re\left[\myv{\mathcal B} e^{i\myv k \cdot \myv r -i\omega t}\right].$$
Where $\myv{\mathcal B}$ comprises 3 complex constants.